Dynamics_Part34

Dynamics_Part34 - V II B and V I C , which is V II B = 1 2...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
where V I A and V I B denote the horizontal velocity components of Spheres A and B just after the impact, respectively. The impact relation tells us that since the coefficient of restitution is e , V I B V I A = e ( V A 0) Thus, rearranging terms, we have the following pair of linear, algebraic equations for V I A and V I B . V I B + V I A = V A V I B V I A = V A e Adding and subtracting the equations yields the solution for V I A and V I B . Noting from Part (a) that V A = g f , the solution is V I A = 1 2 (1 e ) 0 g f and V I B = 1 2 (1 + e ) 0 g f (c) Focusing now on the impact between Sphere B and Sphere C, horizontal momentum is conserved, which yields mV I B +0= mV II B + mV I C where V II B and V I C denote the horizontal velocity components of Spheres B and C just after the impact, respectively. The impact relation tells us that since the coefficient of restitution is e , V I C V II B = e ( V I B 0) Thus, rearranging terms, we have the following pair of linear, algebraic equations for V II B and V I C . V I C + V II B = V I B V I C V II B = V I B e Adding and subtracting the equations yields the solution for
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: V II B and V I C , which is V II B = 1 2 (1 e ) V I B and V I C = 1 2 (1 + e ) V I B Finally, we know from Part (b) that V I B = 1 2 (1 + e ) g f Therefore, the final solution for V II B and V I C is V II B = 1 4 (1 e 2 ) g f and V I C = 1 4 (1 + e ) 2 g f (d) We now have all the information we need to compute the height to which Sphere C rises, f C , and thereby find the value of the coefficient of restitution for the observed value of f C . We know that total mechanical energy is conserved, the initial potential energy of Sphere C is zero and its final kinetic energy is zero. Thus, 1 2 mV I 2 C + 0 = 0 + mg f C = f C = V I 2 C 2 g Substituting for V I C from Part (c), we have f C = (1 + e ) 4 16 g f 2 g = (1 + e ) 4 32 f...
View Full Document

Ask a homework question - tutors are online